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(D) The distance between two consecutive crests is the wavelength $\lambda$. However,if the wave is not necessarily consecutive,the distance between t

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$1, \frac{1}{3}, \frac{1}{5}, \dots$

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(D) The distance between two consecutive crests is the wavelength $\lambda$. However,if the wave is not necessarily consecutive,the distance between two crests is $n_2 \lambda$,where $n_2$ is an integer.Given $n_2 \lambda = 5 \, m \implies \lambda = \frac{5}{n_2}$.The distance between a crest and a trough is given by $(2n_1 + 1) \frac{\lambda}{2}$,where $n_1$ is an integer.Given $(2n_1 + 1) \frac{\lambda}{2} = 1.5 \, m \implies (2n_1 + 1) \lambda = 3 \, m \implies \lambda = \frac{3}{2n_1 + 1}$.Equating the two expressions for $\lambda$: $\frac{5}{n_2} = \frac{3}{2n_1 + 1} \implies 5(2n_1 + 1) = 3n_2 \implies 10n_1 + 5 = 3n_2$.For $n_1 = 1$,$3n_2 = 15 \implies n_2 = 5$,so $\lambda = \frac{5}{5} = 1 \, m$.For $n_1 = 4$,$3n_2 = 45 \implies n_2 = 15$,so $\lambda = \frac{5}{15} = \frac{1}{3} \, m$.For $n_1 = 7$,$3n_2 = 75 \implies n_2 = 25$,so $\lambda = \frac{5}{25} = \frac{1}{5} \, m$.Thus,the possible wavelengths are $1, \frac{1}{3}, \frac{1}{5}, \dots$.

For a transverse wave travelling along a straight line,the distance between two peaks (crests) is $5 \, m$,while the distance between one crest and one trough is $1.5 \, m$. The possible wavelengths (in $m$) of the waves are

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